Find $lim_{ntoinfty}frac{(1+7n+n^3)^4}{3-14n^{12}}$ [closed]
I don't really have any idea how to find $$lim_{ntoinfty}frac{(1+7n+n^3)^4}{3-14n^{12}}$$
Can someone please give me a hint on what to do here? What I'm having trouble with especially is the powers. How do I get rid of them?
Thanks in advance!
real-analysis limits
edited Dec 3 '18 at 13:03
Lorenzo B.
1,8322520
asked Dec 3 '18 at 12:33
D. John
283
closed as off-topic by abiessu, Jyrki Lahtonen, Lord_Farin, user10354138, RRL Dec 3 '18 at 15:05
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – abiessu, Jyrki Lahtonen, Lord_Farin, user10354138, RRL
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
I don't really have any idea how to find $$lim_{ntoinfty}frac{(1+7n+n^3)^4}{3-14n^{12}}$$
Can someone please give me a hint on what to do here? What I'm having trouble with especially is the powers. How do I get rid of them?
Thanks in advance!
real-analysis limits
edited Dec 3 '18 at 13:03
Lorenzo B.
1,8322520
asked Dec 3 '18 at 12:33
D. John
283
closed as off-topic by abiessu, Jyrki Lahtonen, Lord_Farin, user10354138, RRL Dec 3 '18 at 15:05
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – abiessu, Jyrki Lahtonen, Lord_Farin, user10354138, RRL
If this question can be reworded to fit the rules in the help center, please edit the question.
Hint: $(a+bx)^kapprox b^kx^k$ for limits going to infinity.
– abiessu
Dec 3 '18 at 12:37
add a comment |
I don't really have any idea how to find $$lim_{ntoinfty}frac{(1+7n+n^3)^4}{3-14n^{12}}$$
Can someone please give me a hint on what to do here? What I'm having trouble with especially is the powers. How do I get rid of them?
Thanks in advance!
real-analysis limits
edited Dec 3 '18 at 13:03
Lorenzo B.
1,8322520
asked Dec 3 '18 at 12:33
D. John
283
I don't really have any idea how to find $$lim_{ntoinfty}frac{(1+7n+n^3)^4}{3-14n^{12}}$$
Can someone please give me a hint on what to do here? What I'm having trouble with especially is the powers. How do I get rid of them?
Thanks in advance!
real-analysis limits
real-analysis limits
edited Dec 3 '18 at 13:03
Lorenzo B.
1,8322520
asked Dec 3 '18 at 12:33
D. John
283
edited Dec 3 '18 at 13:03
Lorenzo B.
1,8322520
asked Dec 3 '18 at 12:33
D. John
283
edited Dec 3 '18 at 13:03
Lorenzo B.
1,8322520
edited Dec 3 '18 at 13:03
Lorenzo B.
1,8322520
edited Dec 3 '18 at 13:03
Lorenzo B.
1,8322520
1,8322520
asked Dec 3 '18 at 12:33
D. John
283
asked Dec 3 '18 at 12:33
D. John
283
asked Dec 3 '18 at 12:33
D. John
283
283
closed as off-topic by abiessu, Jyrki Lahtonen, Lord_Farin, user10354138, RRL Dec 3 '18 at 15:05
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – abiessu, Jyrki Lahtonen, Lord_Farin, user10354138, RRL
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by abiessu, Jyrki Lahtonen, Lord_Farin, user10354138, RRL Dec 3 '18 at 15:05
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – abiessu, Jyrki Lahtonen, Lord_Farin, user10354138, RRL
If this question can be reworded to fit the rules in the help center, please edit the question.
Hint: $(a+bx)^kapprox b^kx^k$ for limits going to infinity.
– abiessu
Dec 3 '18 at 12:37
add a comment |
Hint: $(a+bx)^kapprox b^kx^k$ for limits going to infinity.
– abiessu
Dec 3 '18 at 12:37
Hint: $(a+bx)^kapprox b^kx^k$ for limits going to infinity.
– abiessu
Dec 3 '18 at 12:37
Hint: $(a+bx)^kapprox b^kx^k$ for limits going to infinity.
– abiessu
Dec 3 '18 at 12:37
add a comment |
5 Answers
5
active
oldest
votes
HINT
$$frac{(1+7n+n^3)^4}{3-14n^{12}}=-frac{left(n^3left(frac1{n^3}+frac7{n^2}+1right)right)^4}{14n^{12}-3}=-frac{n^{12}}{n^{12}} frac{left(frac1{n^3}+frac7{n^2}+1right)^4}{14-frac3{n^{12}}}$$
answered Dec 3 '18 at 13:04
gimusi
1
add a comment |
Hint:
$$frac{(1+7n+n^3)^4}{3-14n^{12}} = frac{(color{blue}{n(n^2+7)}+1)^4}{3-14n^{12}}$$
$(n(n^2+7))^4$ has $n^{12}$ as its leading term. So, you reach something in the form of
$$lim_{x to infty}frac{ax^n+…}{bx^n+…}$$
with $n$ as the highest power. Factoring by $x^n$ results in
$$lim_{x to infty}frac{a+frac{something}{x^{n-1}}+frac{something}{x^{n-2}}+…}{b+frac{something}{x^{n-1}}+frac{something}{x^{n-2}}+…}$$
What do those other terms tend to as $x to infty$?
answered Dec 3 '18 at 12:48
KM101
5,0391423
add a comment |
Hint #1: As $n rightarrow infty$, $1/n rightarrow ?$
Hint #2: How fast does $1/n^k$ approach the answer in Hint #1?
answered Dec 3 '18 at 12:40
bjcolby15
1,1681916
add a comment |
Hint:
- Divide the numerator and the denominator by $n^{12}$ might help.
answered Dec 3 '18 at 12:35
Siong Thye Goh
99.5k1464117
add a comment |
We have $$lim_{ntoinfty}frac{(1+7n+n^3)^4}{3-14n^{12}}{=lim_{ntoinfty}frac{(1+7n+n^3)^4}{-14n^{12}}\=-{1over 14}lim_{ntoinfty}{left({1+7n+n^3over n^3}right)^4}\=-{1over 14}lim_{nto infty}left(1+{7over n^2}+{1over n^3}right)^4\=-{1over 14}lim_{nto infty}left(1+{7over n^2}right)^4\=-{1over 14}}$$
answered Dec 3 '18 at 14:10
Mostafa Ayaz
13.7k3936
add a comment |
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
HINT
$$frac{(1+7n+n^3)^4}{3-14n^{12}}=-frac{left(n^3left(frac1{n^3}+frac7{n^2}+1right)right)^4}{14n^{12}-3}=-frac{n^{12}}{n^{12}} frac{left(frac1{n^3}+frac7{n^2}+1right)^4}{14-frac3{n^{12}}}$$
answered Dec 3 '18 at 13:04
gimusi
1
add a comment |
HINT
$$frac{(1+7n+n^3)^4}{3-14n^{12}}=-frac{left(n^3left(frac1{n^3}+frac7{n^2}+1right)right)^4}{14n^{12}-3}=-frac{n^{12}}{n^{12}} frac{left(frac1{n^3}+frac7{n^2}+1right)^4}{14-frac3{n^{12}}}$$
answered Dec 3 '18 at 13:04
gimusi
1
add a comment |
HINT
$$frac{(1+7n+n^3)^4}{3-14n^{12}}=-frac{left(n^3left(frac1{n^3}+frac7{n^2}+1right)right)^4}{14n^{12}-3}=-frac{n^{12}}{n^{12}} frac{left(frac1{n^3}+frac7{n^2}+1right)^4}{14-frac3{n^{12}}}$$
answered Dec 3 '18 at 13:04
gimusi
1
HINT
$$frac{(1+7n+n^3)^4}{3-14n^{12}}=-frac{left(n^3left(frac1{n^3}+frac7{n^2}+1right)right)^4}{14n^{12}-3}=-frac{n^{12}}{n^{12}} frac{left(frac1{n^3}+frac7{n^2}+1right)^4}{14-frac3{n^{12}}}$$
answered Dec 3 '18 at 13:04
gimusi
1
answered Dec 3 '18 at 13:04
gimusi
1
answered Dec 3 '18 at 13:04
gimusi
1
answered Dec 3 '18 at 13:04
gimusi
1
1
add a comment |
add a comment |
Hint:
$$frac{(1+7n+n^3)^4}{3-14n^{12}} = frac{(color{blue}{n(n^2+7)}+1)^4}{3-14n^{12}}$$
$(n(n^2+7))^4$ has $n^{12}$ as its leading term. So, you reach something in the form of
$$lim_{x to infty}frac{ax^n+…}{bx^n+…}$$
with $n$ as the highest power. Factoring by $x^n$ results in
$$lim_{x to infty}frac{a+frac{something}{x^{n-1}}+frac{something}{x^{n-2}}+…}{b+frac{something}{x^{n-1}}+frac{something}{x^{n-2}}+…}$$
What do those other terms tend to as $x to infty$?
answered Dec 3 '18 at 12:48
KM101
5,0391423
add a comment |
Hint:
$$frac{(1+7n+n^3)^4}{3-14n^{12}} = frac{(color{blue}{n(n^2+7)}+1)^4}{3-14n^{12}}$$
$(n(n^2+7))^4$ has $n^{12}$ as its leading term. So, you reach something in the form of
$$lim_{x to infty}frac{ax^n+…}{bx^n+…}$$
with $n$ as the highest power. Factoring by $x^n$ results in
$$lim_{x to infty}frac{a+frac{something}{x^{n-1}}+frac{something}{x^{n-2}}+…}{b+frac{something}{x^{n-1}}+frac{something}{x^{n-2}}+…}$$
What do those other terms tend to as $x to infty$?
answered Dec 3 '18 at 12:48
KM101
5,0391423
add a comment |
Hint:
$$frac{(1+7n+n^3)^4}{3-14n^{12}} = frac{(color{blue}{n(n^2+7)}+1)^4}{3-14n^{12}}$$
$(n(n^2+7))^4$ has $n^{12}$ as its leading term. So, you reach something in the form of
$$lim_{x to infty}frac{ax^n+…}{bx^n+…}$$
with $n$ as the highest power. Factoring by $x^n$ results in
$$lim_{x to infty}frac{a+frac{something}{x^{n-1}}+frac{something}{x^{n-2}}+…}{b+frac{something}{x^{n-1}}+frac{something}{x^{n-2}}+…}$$
What do those other terms tend to as $x to infty$?
answered Dec 3 '18 at 12:48
KM101
5,0391423
Hint:
$$frac{(1+7n+n^3)^4}{3-14n^{12}} = frac{(color{blue}{n(n^2+7)}+1)^4}{3-14n^{12}}$$
$(n(n^2+7))^4$ has $n^{12}$ as its leading term. So, you reach something in the form of
$$lim_{x to infty}frac{ax^n+…}{bx^n+…}$$
with $n$ as the highest power. Factoring by $x^n$ results in
$$lim_{x to infty}frac{a+frac{something}{x^{n-1}}+frac{something}{x^{n-2}}+…}{b+frac{something}{x^{n-1}}+frac{something}{x^{n-2}}+…}$$
What do those other terms tend to as $x to infty$?
answered Dec 3 '18 at 12:48
KM101
5,0391423
answered Dec 3 '18 at 12:48
KM101
5,0391423
answered Dec 3 '18 at 12:48
KM101
5,0391423
answered Dec 3 '18 at 12:48
KM101
5,0391423
5,0391423
add a comment |
add a comment |
Hint #1: As $n rightarrow infty$, $1/n rightarrow ?$
Hint #2: How fast does $1/n^k$ approach the answer in Hint #1?
answered Dec 3 '18 at 12:40
bjcolby15
1,1681916
add a comment |
Hint #1: As $n rightarrow infty$, $1/n rightarrow ?$
Hint #2: How fast does $1/n^k$ approach the answer in Hint #1?
answered Dec 3 '18 at 12:40
bjcolby15
1,1681916
add a comment |
Hint #1: As $n rightarrow infty$, $1/n rightarrow ?$
Hint #2: How fast does $1/n^k$ approach the answer in Hint #1?
answered Dec 3 '18 at 12:40
bjcolby15
1,1681916
Hint #1: As $n rightarrow infty$, $1/n rightarrow ?$
Hint #2: How fast does $1/n^k$ approach the answer in Hint #1?
answered Dec 3 '18 at 12:40
bjcolby15
1,1681916
edited Dec 5 '18 at 12:09
answered Dec 3 '18 at 12:40
bjcolby15
1,1681916
answered Dec 3 '18 at 12:40
bjcolby15
1,1681916
answered Dec 3 '18 at 12:40
bjcolby15
1,1681916
1,1681916
add a comment |
add a comment |
Hint:
- Divide the numerator and the denominator by $n^{12}$ might help.
answered Dec 3 '18 at 12:35
Siong Thye Goh
99.5k1464117
add a comment |
Hint:
- Divide the numerator and the denominator by $n^{12}$ might help.
answered Dec 3 '18 at 12:35
Siong Thye Goh
99.5k1464117
add a comment |
Hint:
- Divide the numerator and the denominator by $n^{12}$ might help.
answered Dec 3 '18 at 12:35
Siong Thye Goh
99.5k1464117
Hint:
- Divide the numerator and the denominator by $n^{12}$ might help.
answered Dec 3 '18 at 12:35
Siong Thye Goh
99.5k1464117
answered Dec 3 '18 at 12:35
Siong Thye Goh
99.5k1464117
answered Dec 3 '18 at 12:35
Siong Thye Goh
99.5k1464117
answered Dec 3 '18 at 12:35
Siong Thye Goh
99.5k1464117
99.5k1464117
add a comment |
add a comment |
We have $$lim_{ntoinfty}frac{(1+7n+n^3)^4}{3-14n^{12}}{=lim_{ntoinfty}frac{(1+7n+n^3)^4}{-14n^{12}}\=-{1over 14}lim_{ntoinfty}{left({1+7n+n^3over n^3}right)^4}\=-{1over 14}lim_{nto infty}left(1+{7over n^2}+{1over n^3}right)^4\=-{1over 14}lim_{nto infty}left(1+{7over n^2}right)^4\=-{1over 14}}$$
answered Dec 3 '18 at 14:10
Mostafa Ayaz
13.7k3936
add a comment |
We have $$lim_{ntoinfty}frac{(1+7n+n^3)^4}{3-14n^{12}}{=lim_{ntoinfty}frac{(1+7n+n^3)^4}{-14n^{12}}\=-{1over 14}lim_{ntoinfty}{left({1+7n+n^3over n^3}right)^4}\=-{1over 14}lim_{nto infty}left(1+{7over n^2}+{1over n^3}right)^4\=-{1over 14}lim_{nto infty}left(1+{7over n^2}right)^4\=-{1over 14}}$$
answered Dec 3 '18 at 14:10
Mostafa Ayaz
13.7k3936
add a comment |
We have $$lim_{ntoinfty}frac{(1+7n+n^3)^4}{3-14n^{12}}{=lim_{ntoinfty}frac{(1+7n+n^3)^4}{-14n^{12}}\=-{1over 14}lim_{ntoinfty}{left({1+7n+n^3over n^3}right)^4}\=-{1over 14}lim_{nto infty}left(1+{7over n^2}+{1over n^3}right)^4\=-{1over 14}lim_{nto infty}left(1+{7over n^2}right)^4\=-{1over 14}}$$
answered Dec 3 '18 at 14:10
Mostafa Ayaz
13.7k3936
We have $$lim_{ntoinfty}frac{(1+7n+n^3)^4}{3-14n^{12}}{=lim_{ntoinfty}frac{(1+7n+n^3)^4}{-14n^{12}}\=-{1over 14}lim_{ntoinfty}{left({1+7n+n^3over n^3}right)^4}\=-{1over 14}lim_{nto infty}left(1+{7over n^2}+{1over n^3}right)^4\=-{1over 14}lim_{nto infty}left(1+{7over n^2}right)^4\=-{1over 14}}$$
answered Dec 3 '18 at 14:10
Mostafa Ayaz
13.7k3936
answered Dec 3 '18 at 14:10
Mostafa Ayaz
13.7k3936
answered Dec 3 '18 at 14:10
Mostafa Ayaz
13.7k3936
answered Dec 3 '18 at 14:10
Mostafa Ayaz
13.7k3936
13.7k3936
add a comment |
add a comment |
Hint: $(a+bx)^kapprox b^kx^k$ for limits going to infinity.
– abiessu
Dec 3 '18 at 12:37